【Pat】甲级1004 - Counting Leaves（树 & bfs）

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1004.Counting Leaves (30)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input

Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.
Output

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output “0 1” in a line.

Sample Input
2 1
01 1 02
Sample Output
0 1

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存树的时候用vector来存边，然后从根节点开始bfs遍历就行了。

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代码如下：

#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<queue>
#include<stack>
using namespace std;
typedef long long LL;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define fi first
#define se second
vector<int> Node[105];
void bfs(int root)
{
    queue<pair<int,int> > Q;
    Q.push(make_pair(root,1));
    int cnt = 0;
    int pos = 1;
    while (!Q.empty())
    {
        if (pos != Q.front().se)        //下一层
        {
            cout << cnt << ' ';     //输出上一层的数量
            cnt = 0;
            pos++; 
        }
        if (Node[Q.front().fi].size() == 0)     //没有孩子，那么它为叶子 
            cnt++;
        for (int i = 0 ; i < Node[Q.front().fi].size() ; i++)       //搜索它的孩子 
            Q.push(make_pair(Node[Q.front().fi][i] , pos+1));
        Q.pop();        //最后把这个节点弹出队列 
    }
    cout << cnt << endl;        //输出最后一层的数量 
}
int main()
{
    int id;
    int in[105] = {-1};
    int n,m,k;
    cin >> n >> m;
    while (m--)
    {
        cin >> id >> k;
        if (in[id] == -1)
            in[id] = 0;
        while (k--)
        {
            int t;
            cin >> t;
            Node[id].push_back(t);
            if (in[t] == -1)
                in[t] = 0;
            in[t]++;
        }
    }
    int root;
    for (int i = 0 ; i <= 99 ; i++)
    {
        if (!in[i])     //入度为0
        {
            root = i;
            break;
        }
    }
    bfs(root);
    return 0;
}